/* Sample solution to Watchdog
 * Author: Mikael Goldmann
 *
 * Try all possible points of attachment
 */


import java.io.*;
import java.util.*;

class dog_mg
{
    static BufferedReader stdin;
    static PrintStream stdout;
    
    public static void main(String[] a) throws Exception
    {
	Reader rdr = new InputStreamReader(System.in);
	stdin = new BufferedReader(rdr);
	OutputStream os = new BufferedOutputStream(System.out);
	stdout =  new PrintStream(os);
	
	(new dog_mg()).run();
	
	stdout.close();
	if (stdin.readLine() != null)
	    System.err.println("*** Stuff after last test case ***");
    }

    void run() throws Exception
    {
	int n=0;
	try {
	    n =	Integer.parseInt(stdin.readLine());
	}
	catch(Exception e) {
	    System.err.println("*** Cannot read number of cases ***");
	    throw e;
	}	    
	for (int i = 1; i <= n; ++i) {
	    try {
		solve(i);
	    }
	    catch(Exception e) {
		System.err.println("***Problem in case " + i + " ***");
		throw e;
	    }
	}
    }
 
    void solve(int cnt) throws Exception
    {
	String s = stdin.readLine();
	StringTokenizer st = new StringTokenizer(s);
	int S = Integer.parseInt(st.nextToken());
	int H = Integer.parseInt(st.nextToken());
	int i;
	int[] hx = new int[H];
	int[] hy = new int[H];
	for (i=0; i < H; ++i) {
	    s = stdin.readLine();
	    st = new StringTokenizer(s);
	    hx[i] = Integer.parseInt(st.nextToken());
	    hy[i] = Integer.parseInt(st.nextToken());
	}
	int x,y;
	int best = 10000;
	for (x=1; x < S; ++x)
	    for (y=1; y < S; ++y) {
		boolean onHatch = false;
		for (i=0; !onHatch && i < H; ++i)
		    onHatch = (x == hx[i] && y == hy[i]);
		if (!onHatch) {
		    int leash = 
			Math.min(Math.min(x, y), Math.min(S-x, S-y));
		    boolean all = true;
		    for (i=0; all && i < H; ++i) {
			int dx = x - hx[i];
			int dy = y - hy[i];
			all = dx*dx + dy*dy <= leash*leash;
		    }
		    if(all) { // Solution found!
			stdout.println(x + " " + y);
			return;
		    }
		}
	    }
	stdout.println("poodle");
    }
}